Question

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here's your question ⏩⏩
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$<b><u>Prove that:-</u></b>$

$\bold {s = \frac{n}{2} [2a + (n - 1)d] }$
where a = 1st term,d =common difference

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Answer 1

hey mate.
here's the solution

Answer 2

In AS,

a = first term

d = common difference

n = number of terms

l = last term or

l th term = a + ( n - 1 )d

We know that $S_{n}$ is the sum of all the terms. In the question, let there are n terms. So

$S_{n}=a+(a+d)+(a+2d)+(a+3d)+.........(l-3d)+(l-2d)+(l-d)+l$

$S_{n}=a+a+d+a+2d+a+3d+.........l-3d+l-2d+l-d+l\:\:\:\:\:\:\:\:\:\:\:\:\:\textit{\textbf{...(i)}}$

The sum of n terms can also be written in the reverse form of this.

$S_{n}=l+(l-d)+(l-2d)+(l-3d)+......(a+3d)+(a+2d)+(a+d)+a$

$S_{n}=l+l-d+l-2d+l-3d+......a+3d+a+2d+a+d+a\:\:\:\:\:\:\:\:\:\:\:\:\textit{\textbf{...(ii)}}$

     

Adding ( i ) & ( ii ) ,

$2S_{n}=a+l+a+d+l-d+a+2d+l-2d+a+3d+l-3d+.......a+3d+l-3d+a+2d+l-2d+a+d+l-d+a+d$

$2S_{n}=(a+l)+(a+l)+(a+l)+.........n\:terms$

$2S_{n}$=n( a + l )

$S_{n}=\dfrac{n}{2}(a+l)$

l th term = a + ( n - 1 )d

$S_{n}=\dfrac{n}{2}(a+a+(n-1)d)$

$S_{n}=\dfrac{n}{2}(2a+(n-1)d$

                Proved